Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $a = \dfrac{-3z + 9}{z^2 + z - 90} \div \dfrac{3z + 15}{z^2 + 15z + 50} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $a = \dfrac{-3z + 9}{z^2 + z - 90} \times \dfrac{z^2 + 15z + 50}{3z + 15} $ First factor out any common factors. $a = \dfrac{-3(z - 3)}{z^2 + z - 90} \times \dfrac{z^2 + 15z + 50}{3(z + 5)} $ Then factor the quadratic expressions. $a = \dfrac {-3(z - 3)} {(z + 10)(z - 9)} \times \dfrac {(z + 10)(z + 5)} {3(z + 5)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac {-3(z - 3) \times (z + 10)(z + 5) } { (z + 10)(z - 9) \times 3(z + 5)} $ $a = \dfrac {-3(z + 10)(z + 5)(z - 3)} {3(z + 10)(z - 9)(z + 5)} $ Notice that $(z + 10)$ and $(z + 5)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac {-3\cancel{(z + 10)}(z + 5)(z - 3)} {3\cancel{(z + 10)}(z - 9)(z + 5)} $ We are dividing by $z + 10$ , so $z + 10 \neq 0$ Therefore, $z \neq -10$ $a = \dfrac {-3\cancel{(z + 10)}\cancel{(z + 5)}(z - 3)} {3\cancel{(z + 10)}(z - 9)\cancel{(z + 5)}} $ We are dividing by $z + 5$ , so $z + 5 \neq 0$ Therefore, $z \neq -5$ $a = \dfrac {-3(z - 3)} {3(z - 9)} $ $ a = \dfrac{-(z - 3)}{z - 9}; z \neq -10; z \neq -5 $